How to replace CRLF if not end of line using sed on windows or powershell

Scripting/Programming
,
1

I have a large text file with millions of rows on windows. Some of the records have CRLF in the middle of the record (description field which is free hand text). So, I end up with lines that looks like this…

xxxxxxxxxxxxxxxCRLFxxxxxxxxxxxxxxxxxxCRLF
xxxxxxxxxxxxxxCRLF
xxxxxxxxxxxxxxxxCRLFxxxxxxxxxxxxxCRLF

So, I need a script that removes CRLF if not at the end of line. Powershell or sed ideal.
Thanks in advance.

2

take a look at the following page
https://www.cyberciti.biz/faq/sed-remove-m-and-line-feeds-under-unix-linux-bsd-appleosx/

3

Thanks. I have read this before and did not get what I am looking for.

4

Here is a sample of what I find like record 237:
208|7445|6|1|16933|16941| |10963| |1PA0634064|2016-03-10 00:00:00.000|2016-09-10 00:00:00.000|0||1|16945|1|16947|1|16960|2016-03-17 15:40:08.000|363|2016-03-17 15:40:34.000|363|0||0
236|9029|4|1|16934|16940| |10964| |33-XX9029-07-1001-01-APP|2016-04-01 00:00:00.000|2017-04-01 00:00:00.000|0||0||0||0||2016-03-22 09:24:28.000|647|2016-04-04 14:22:57.000|647|1|16947|1
237|9029|4|1|16933|16941| |10963|on file

| |||0||1|16945|1|16947|1|16960|2016-03-22 09:30:34.000|647|||0||0
238|7709|10|1|16934|16940| |10964| |33-011777-07-1001-02-APP|2016-04-20 00:00:00.000|2017-04-20 00:00:00.000|0||0||0||0||2016-03-22 11:42:55.000|363|||1|16947|1
244|7479|6|1|16933|16941| |10963| |1PA3688562|2015-05-20 00:00:00.000|2016-05-20 00:00:00.000|0||1|16945|1|16947|1|16980|2016-03-22 13:19:30.000|363|||0||0
248|7157|4|1|16939|16940| |10963| |32-105946-14-004-6-02|2015-08-28 00:00:00.000|2016-08-28 00:00:00.000|0||0||0||0||2016-03-22 13:32:53.000|363|||1|16946|0